Chapter 38, Problem 045 Assuming that your surface temperature is 97.1 F and tha

Question

Chapter 38, Problem 045 Assuming that your surface temperature is 97.1 F and that you are an ideal blackbody radiator (you are close), find (a) the wavelength at which your spectral radiancy is maximum, (b) the power at which you emit thermal radiation in a wavelength range of 1.10 nm at that wavelength, from a surface area of 4.10 cm2, and (c) the corresponding rate at which you emit photons from that area. Using a wavelength of 500 nm (in the visible range), (d) recalculate the power and (e) the rate of photon emission. (As you have noticed, you do not visibly glow in the dark.) (a) Number (b) Number (c) Number (d) Number (e) Number Units Units Units Units Units

Explanation / Answer

part a )

T = 97.1 F = 36.17 C = 309.32 K

use wien law

lambda_max = 2898 * um .K/T = 2898/309.32 = 9.37 um = 9.37 x 10^-6 m

part b )

lambda = 9.37 um , T = 309.32 K

spectral radiancy = s = (2*pi*c^2*h/lambda^5) * ( 1/e^hc/lambda*k*T -1)

k =boltzman constant

s = 3.58 x 10^7 W/m^3

For small range of wavelength, the radiated power

P = s*A*(dlambda)

dlambda = 1.1 x 10^-9 m

A = 4.10 cm^2 = 4.10 x 10^-4

P = 1.61 x 10^-5 W

part c )

E = hc/lambda

lambda = 9.37 x 10^-6 m

E = 2.12 x 10^-20 J

dN/dt = P/E = 7.58 x 10^14 photon/s

part d )

spectral radiancy = s = (2*pi*c^2*h/lambda^5) * ( 1/e^hc/lambda*k*T -1)

lambda = 490 nm = 500 x 10^-9 m

s = 6.255 x 10^-26 W/m^3

P = s*A*dlambda

P = 2.82 x 10^-36 W

part e )

E = hc/lambda

dN/dt = P/E = 7.09 x 10^-18 photon/s

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.